Question

i Question. You are working for a Biotechnology company and with ghes you a tak to help create a viewem odcing a Eastoretreatshown in the action low. You have alre Rhod o dectes that performs the reaction so you decide to the whole cells n a n your d foran d the mu d You fit by the same name a s the difiche The With mixed luck from the wyme, you try using the whole Al shows the Where would have the po productivity of the cells for making t h e w the er marks) c tie 6. Would the harvest time change if you wanted to colle wows) d e Ahogh you have a good Further resourc from the c o e is being While you will bring to work with the to the profilului Although the destu the Youndthew's l bych

Answer 1

hug off at a concentration Ansi- a) the reaction is levelling off at a con of - 55-60 mg/L of the substrate. [5]. The Enzyme must be saturated with the substral at this concentration and thus levelling off product formation Firstly, this would suggest that the [5] is in Excess ise much higher than the km. According to Michaelis a Menten Kinetics, km is the Concentration of the substrate at which the rate of reaction is half of its maximun velocity (Vmar). Km denotes the affinity of the substrate to the Euryure. when the reaction has reached Imax I any farther addition of [s] will not impact the rate of reaction ise the rate at which the substrate is turned over into the product. for a reaction to proceed at an optimal rate you always want to supply the [sjat a concentration that is at least 5x10x >> KM. Decrease the [s] to graphically abserve complete sonover. Secondly, if you want to slow down the rate of this reaction ise if you want to expand the linear range over a larger period of time

then you could consider decreasing the concentra. tion of the Euzyne. This can be achieved by diluting the your Enzywe. b) The product appears to be produced in tandem with cell growth ise at the start of the exponential phase. To obtain maximum product yield, you would harvest the product just before the stationary phase of growtte i, e after in 220 minutes . The average rate of Hoppen production u toomg/L / 220 minuter = 0.45 mg /4/ min c) Primary metabolites can be collected from the growth phase of the organisme je between 70 - 200 minutes approximately. dy the intracellular réphoning of the product can be genetically overcome by knocking out the genes Encoding the Euryme that is consuming the product. Alternatively, you can manipulate the reaction by including a inhibitor of the (second Enzepone) in your reaction mixture

e) According to the lock & key hypother's, the substrate fits into the catalytic rête of the Enzyme in a lock as key orientation ise et is a Perfect fit. This is possible mainly due to strong covalent bond interactions in the catalytic site. The actuity of the Euregimes, with mutations in Glycine. Imp & Protire residues suggest that the affinity of the substrate to the Enzyme has changed as a result of a change in structural orientation site directed mutagenesis to glycine has resulted in a favourable increase in activity where as the Tespa Pro mutations have greatly decreased the Enzyme acturity indicating a proof fit of the substrate into the catalytie site of Enzene. Pleast rate my Answer.